I try to classify non-abelian groups of order $56$ with sylow $2$-subgroup isomorphic to quaterion group $Q_8$. More accurately I want to construct $2$ non-isomorphic such groups. This is an excercise 5.3.7 from Dummit and Foote's book. I can construct such groups as semidirect product as follows $G_1 = C_7 \rtimes_{\phi_{1}} Q_8$, $G_2 = C_7 \rtimes_{\phi_{2}} Q_8$, $G_3 = C_7 \rtimes_{\phi_{3}} Q_8$, where $\phi_n: Q_8 \to \operatorname{Aut}(C_7)$. Let $Q_8 = <i,j>$ and $<\sigma>$ be a unique subgroup of order $2$ of $\operatorname{Aut}(C_7) \cong C_6$, where $\sigma$ inverts elements of $C_7$. We define $\phi_1$ as $$\phi_1(i) = \sigma, \phi_1(j) = \operatorname{id}$$ (here $\operatorname{id}$ is an identical automorphism), $$\phi_2(i) = \operatorname{id}, \phi_2(j) = \sigma$$ and $$\phi_3(i) = \sigma, \phi_3(j) = \sigma$$

My question. Am I right that all this groups are isomorphic? And if all this groups are isomorphic what is the second type of such gruops?

I use the following result about isomorphism of semidirect product. Let $\tau\in \operatorname{Aut}(H)$ than $N\rtimes_{\phi} H \cong N\rtimes_{\tau\phi} H$.


You are correct: those are all isomorphic. The second type is of course the direct product. That is different as $C_7$ is central in the direct product of $C_7$ with $Q_8$.

  • $\begingroup$ Thank you very much! For some reason I do not know why I thought that we should construct precisely non-trivial homomorphisms =) $\endgroup$ – Mikhail Goltvanitsa Dec 13 '18 at 18:24

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